Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
inline char gc() {
  static const int L = 1000;
  static char sxd[L], *sss = sxd, *ttt = sxd;
  if (sss == ttt) {
    ttt = (sss = sxd) + fread(sxd, 1, L, stdin);
    if (sss == ttt) return EOF;
  }
  return *sss++;
}
inline int read(int& x) {
  x = 0;
  char c = getchar();
  bool f = false;
  for (; !isdigit(c); c = getchar()) {
    if (c == EOF) return -1;
    if (c == '-') f = true;
  }
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return 1;
}
int i, j, f, n, m, g[1000][1000], a[1000][1000], b[1000], c[1000], num;
int main() {
  read(n);
  read(m);
  for (i = 0; i < n; i++) b[i] = 1000;
  for (i = 0; i < m; i++) c[i] = 1000;
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) read(g[i][j]);
  if (n <= m) {
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) b[i] = min(b[i], g[i][j]);
      for (j = 0; j < m; j++) g[i][j] -= b[i];
      num += b[i];
    }
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) c[i] = min(c[i], g[j][i]);
      for (j = 0; j < n; j++) g[j][i] -= c[i];
      num += c[i];
    }
  } else {
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) c[i] = min(c[i], g[j][i]);
      for (j = 0; j < n; j++) g[j][i] -= c[i];
      num += c[i];
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) b[i] = min(b[i], g[i][j]);
      for (j = 0; j < m; j++) g[i][j] -= b[i];
      num += b[i];
    }
  }
  f = 1;
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++)
      if (g[i][j]) {
        f = 0;
        break;
      }
  if (f) {
    cout << num << endl;
    for (i = 0; i < n; i++)
      for (j = 0; j < b[i]; j++) printf("row %d\n", i + 1);
    for (i = 0; i < m; i++)
      for (j = 0; j < c[i]; j++) printf("col %d\n", i + 1);
  } else
    puts("-1");
  return 0;
}